Fiber Optics Worksheet
Name:
The use of optical fibers to transmit information became widespread in the late 1980s, when traffic systems
and phone networks began to use the technology. Nowadays, high definition cable, 4G phone networks and
even medical imaging use fiber optics heavily. You are going to look at how the improvements in technology
over the years have made the cost more affordable.
First we need to review the basics. When light travels inside a fiber, some light is always lost. Whether it be
due to bending of the fiber, impurities in the material or other physical processes, we will never get as much
light out of the fiber as we put in. The equation for how much light we will get out of the fiber compared to
how much is put in is:
𝑷(𝒙) = 𝑷𝟎 × 𝟏𝟎−𝜶𝒙
In this equation:
P0 is the INPUT POWER, generally given in units of watts.
P(x) is the OUTPUT POWER, also in units of watts.
α is the LOSS COEFFICIENT, and is in units of inverse meters
x is the LENGTH of the fiber, which is usually in units of meters.
1a. If we put 2 Watts of power into a fiber with a loss coefficient of α = 0.5 per kilometer and the fiber is 8
kilometers long, what will the output power be?
1b. Assuming P0 = 2W and α = 0.5 per kilometer, plot P(x) as a function of x. What is the domain and range
given our situation?
x - kilometers
0
.5
1
P(x) – Watts
1.5
2
2.5
3
Domain:
Range:
2. It is often more useful to look at the % of power we are getting out of the fiber compared to what we put in.
a.Calculate from your results for 1a:
% 𝑜𝑓 𝑝𝑜𝑤𝑒𝑟 𝑟𝑒𝑡𝑎𝑖𝑛𝑒𝑑 =
𝑜𝑢𝑡𝑝𝑢𝑡 𝑝𝑜𝑤𝑒𝑟
=
𝑖𝑛𝑝𝑢𝑡 𝑝𝑜𝑤𝑒𝑟
b. It will be useful from here on out to look at % of power retained as a function instead of output power.
Define the % of power retained as R(x). Write the equation in the top half of the first page in terms of R(x), α
and x. We will use this equation for the rest of the worksheet.
c. Assuming α = 0.5 per kilometer, plot R(x) as a function of x. What is the domain and range given our
situation?
x - kilometers
0
0.5
1
R(x)
1.5
2
2.5
3
Domain:
Range:
What is the difference between this graph and the graph in 1b?
3. It happens that when R(x) gets below 1%, we can no longer recover the information we input into the
fibers – clearly we cannot allow this to happen. Every time the output power drops to 1% of the input power,
we will need to put in an optical regenerator to boost the power to its original value.
a. Now lets’ look at a different system. In this system, we only get 1% (so R(x) = 0.01) of the power out of the
system that we put in. If α is 0.5 per kilometer, what is x?
b. Explain in words what the value of x you found in part a means in regard to the need for an optical
generator
c. If we needed the fiber in #3 to go 10 kilometers instead of 4 kilometer, how many optical regenerators
would we need?
d. Accounting for the optical regenerators, plot R(x) vs x for α = 0.5 per kilometer. Pay careful attention to
what happens at x=4 & x=8
x - kilometers
0
2
4
R(x)
6
8
10
5. In 1966, two researchers in England decided to use optical fibers in their building. At that point as the fiber
technology was completely new, the loss coefficient was very high - 100 per kilometer.
a. How long would the fiber have to be for R to be 1%?
b. Let’s say they wanted to use optical fibers to transmit information over the length of the U of A’s campus,
which is about a mile or 1.6 kilometers. About how many optical regenerators would they need to use?
c. Let’s say they wanted to cover the US from coast to coast in optical fibers – about 5,000 kilometers. How
many regenerators would be needed?
6. By 1970, major advances had been made. By improving the production process, the company Corning Glass
based in Rochester was able to get α down to 2 per kilometer.
a. How long would the fiber have to be for R to be 1%?
b. Let’s say they wanted to use optical fibers to transmit information over the length of the U of A’s campus,
which is about a mile or 1.6 kilometers. About how many optical regenerators would they need to use?
c. Lets say they wanted to cover the US from coast to coast in optical fibers – about 5,000 kilometers. How
many regenerators would be needed?
7. By the late 1980s, the value of α for an optical fiber was 0.02 per kilometer.
a. How long would the fiber have to be for R to be 1%?
b. Let’s say they wanted to use optical fibers to transmit information over the length of the U of A’s campus,
which is about a mile or 1.6 kilometers. About how many optical regenerators would they need to use?
c. Lets say they wanted to cover the US from coast to coast in optical fibers – about 5,000 kilometers. How
many regenerators would be needed?
8. Now we will derive a cost model for implementing an optical network. Assume α = 0.02 per kilometer
a. How long would the fiber have to be for R to be 1%?
b. Write a function G(N) for the total number of regeneratorsneeded. It will be a function of the total network
length N that we want to use. If you are confused, look back at 7b and 7c and think about what number
represents the network length in that problem, and replace it with N
c. A quick check with Verizon shows it costs $60,000 per kilometer to lay fiber on land (they have to put it far
underground) and $3,000 per regenerator. Make a function for the total cost C(G,N) which is a function of the
number of generators G and total network length N. Replace G with the expression you found in b so that
your result is only a function of N.
d. There is currently a 250,000 km intercontinental undersea fiber network that is undersea. Assuming the
prices in part b, and α=0.02 per km, how much did this network cost?
e. Is assuming the prices in part c are applicable to the network in part d reasonable? Why or why not?